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On the Determination of a Tridiagonal Matrix from Its Spectrum and a Submatrix

P. Deift*

Courant Znstitute of Mathematical Sciences

New York University

New York, New York 10012

and

T. Nanda’

University of California, Berkeley

Lawrence Berkeley Laboratory

Berkeley, California 9r1720

Submitted by Gene H. C.&b

ABSTRACT

Given a set of 2n real numbers A, < X, < . . . < A,, , the authors describe the set { S } of n X n tridiagonal matrices with the property that each S can be completed to a 2n x2n tridiagonal matrix L with spec(L)= {X,,A,,...,X,,}.

INTRODUCTION

In this paper we consider a question in inverse matrix theory raised by G. Golub. Suppose that an n x n matrix

I a1 b, 0 0 0 0 ... 0

b, a2 b, 0 0 0 ... 0

S=O b,a,b, 0 0 ... 0 .& . . b. . .d. . ‘o’ . . . . . . . . . .b 1 . . .;.._,. .

n-2

‘b”’ nl

\ o o o 0 .-- 0 bnpl a, I

*Reseamh sponsored by NSF grant #MCS-8002561.

‘This work was supported in part by the Director, Office of Energy Research, Office of Basic Energy Sciences, Engineering, Mathematical and Geosciences Division of the U.S. Depart- ment of Energy, under contract W-7404-ENG48.

LINEAR ALGEBRA AND ITS APPLZCATZONS 60:43-55 (1984) 43

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44 P. DEIFT AND T. NANDA

( bi > 0) is given together with a set of 2n real numbers A r < A, < . . . < A,,. Can S be completed to a tridiagonal2n X 2n matrix

L=

‘a, b, 0 0 0 ..’ 0 0 ... 0 0

b, a2 b, 0 0 ... 0 0 ... 0 0

0 b, a3 b, 0 ... 0 0 . . . 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

0 0 ... 0 bn_2 a,_, b,_, 0 0 . . . 0

0 0 ... 0 0 bn_l a,, b,, 0 . . . 0

0 0 ... 0 0 0 b,, a,+, b ... n+l 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

0 0 ... 0 0 0 . . . 0 h-2 a+1 h-l (0 0 ... 0 0 0 ... 0 0 b-1 a2n

(bi > 0) with spectrum A,, A,,..., A,,? Hochstadt [l] ’ has shown that if such an L exists, it is unique. In this paper we characterize the space 9of matrices S which can be completed to an L with fixed spectrum A, < A, < . . . < A,,. In particular, we show that Y is the section of a cone by a hyperplane.

The continuum version of this problem is concerned with the following practical question: given a violin string of variable density in 0 < x Q L, can the string be extended to 0 Q x < 2 L and vibrate with given tones X r < X 2 < A,<...?

Lemmas 1, 2, and 3 below as well as Theorem 1 describe standard results (see e.g. [2, 31) from the inverse theory for Jacobi matrices and are included for the reader’s convenience. Theorem 2 is the main result of the paper.

Let L be an n X n real symmetric tridiagonal matrix with

Lii = Lji' 1 G i, j < n,

L,, = 0 if j>i+l,

Liitl z=- 0.

Let U denote the orthogonal matrix consisting of the normalized eigenvectors of L, and let A be the diagonal matrix consisting of the eigenvalues of L, so that

L = UAU’. (1)

‘See ah [31, [41, I51, [61.

DETERMINATION OF A TRIDIAGONAL MATRIX 45

Here U” stands for the transpose of U. From Equation (1) we get the following basic formulae, which will be used in the sequel:

Lkj=(eL,Lej)=(U’e,,AU’ej),

where

e,=(O,O ,..., 1,O ,..., 0)’ (3)

and the 1 occurs in the kth slot. Under the stated assumptions it is easy to show that the eigenvalues of L

are all distinct and the first component of each eigenvector is # 0. Let Lu, = Xkuk, so that

and define the vector f by

f = U’e,, (4)

i.e., f is the first row of U and consists of the first components of the eigenvectors of L. Since the eigenvectors are defined up to a + sign, we adopt a convention that fi are all positive. The first lemma gives an elemen- tary but basic result.

LEMMA 1.

(a) Let L be a real symmetric n X n tridiagonal matrix. For 2 < k < n - 2, Lk is a band matrix with

(Lk)ij=O for Ii- jl> k.

(b) IfLii+l#Oforeveryi,then(Lk)ii+kfOfori<n-k,2<k<n-1.

cc) If (Lk),k+, +Oforsomekintheset{2,3,...,n-l},thenLii+,+O for 1 < i < k.

Proof. (By induction.) (a): Let j = i + p and 2 < k < n - 2. Then

= Lii_,(Lk-‘)i-li+p + Lii(Lk-‘)ii+, + Lii+l(Lkml)i+li+,.


Ifp>k,theni+p-(i-l)>k-1andi+p-(i+1)>k-1.Hence(Lk),j = 0 if the result is assumed to be true for k - 1. Since the result holds for k = 1, (a) follows by induction and the symmetry of L.

(b):

=Lii+l(Lk-l)i+li+k by (41 y

and the result follows by induction.

(c):

(Lk) lk+l = L,2(Lk-‘)2k+,

= L,2L,(Lk-2h3k+l

= L12L23L,, . . . Lkk+l by (41 7

and (c) follows.

Let ok = U ‘e,. Observe that v1 = f.

COROLLARY. (~kf,Vk+2+j)=Ofbd~j~~----, l<k<n-1.

Proof. Using (2), we have

LEMMA 2. Let L be a real symmetric n x n tridiagonul matrix. Then

Lii+l # 0 for every i, 1 < i < n - 1, implies that the n vectors

f, Af, A2f>...> A”- ‘fare linearly independent.

REMARK. As det[ f, A f, . . . , An-‘f]=(n~~=,~)VIAl,A, ,..., X,], where V is the Vandermonde determinant, we see that (in the case that all the eigenvalues are distinct) f, A f, . . . , A”- ‘fare linearly independent if and only if x # 0 for 1~ i < n. We will use this fact often and without further comment.

Proof. Suppose that for some real numbers al, (Ye,. . . , a,

qf+a2Af+ ... +a,A”-‘f=o. (*>


Taking the inner product with II, gives (in view of the corollary above)

a,( A”_‘f, we”) = a ( L”-ql” = 0. ”

However, by Lemma l(b) (L”- l)ln # 0 so (Y, = 0. Next taking the inner product in ( * ) with U, _ 1 gives (Y, _ r = 0. Continuing this way finally gives (Y1 = 0. n

LEMMA 3. Let L be a real symmetric n x n tridiagonul matrix. lf f, Af,..., A"-'f are linearly independent, then the vectors vl, v2,. . . ,v, (which are the rows of the matrix U) are obtained by the Gram-Schmidt orthogonal- izution process applied to these n vectors, and Li i + r f 0 for 16 i < n - 1.

Proof. Let [ * . * ] denote the vector space spanned by the entries in the square bracket. Since the matrix U is orthogonal, vr, vs, . . . , v, are orthonormal vectors. Now

b21= [ “l,v~,v~,...,vn] L

= hl L NfAfl

(by the corollary to Lemma 1); here I denotes the orthogonal complement. Consequently vs is obtained by orthogonalizing f and A f relative to f, and we can write

02 = af + PAf,

where (Y and p are chosen so that (va, f) = 0 and ( v2, v2) = 1. Since ( v2, v2)

= P(Af> 02) = P-&2> it follows that L,, f 0. We also note that v2 can be normalized in a unique way if LIZ > 0.

Next,

[%I = [ “1,V,,V,,...,V,3 L

i.e.,

hl= [fJfl 1 n[fAfA2f].


In particular, 1 = (us, 0s) = const (us, A’f) = const (L’),,, and hence (L’),, # 0, which by Lemma l(c) implies that L, f 0. More generally one has

bkl = [f, Af,..., Ak-“f] ’ n[f,Rf ,..., Ak-If],

and 1 = (v,, uk) = const(u,, Akklf) = const(lk-‘)ik gives (Lk-‘)ik # 0, which by (c) of Lemma 1 implies that L, k+ 1 # 0. This completes the proof of the lemma. W

REMARKS.

(1) Note that if we are given an arbitrary unit vector f and an arbitrary real diagonal matrix A such that f, A f, . . . , A”- ‘f are linearly independent and if v1,v2,..., u, are constructed by applying the Gram-Schmidt orthog- onahzation process to f, A f, . . . , A”- 'f, then

(A~~,u,)=0 for lj-kl>l.

To see this, observe that if k = j + p with p 2 2, then ok belongs to [f, Af,...,Ak-2fli [Af, A2f,...,Aifl,

and vi belongs to [f, A f,. . . , Aj-'f 1, so Avj belongs to and since k - 2 > j, (u,, Auj) = 0. This remark will be

used in Theorem 1 below. (2) In carrying out the Gram-Schmidt process each ok is determined up to

a k sign. However, if Li i+ i > 0 for 1 < i < n - 1, then ( Lk)i k+ i > 0, so that the sign of vk+ 1 is fixed and the vectors ui, 02,. . . , v, are determined uniquely.

Let9(hi,A2,..., h,) denote the space of ah real symmetric n X n tridiag- onaI matrices L with fixed spectrum A, -C X2 <. . . < A, and Lii+l > 0 for l<i<n-1.

THEOREM 1. Let L be a real symmetric n X n tridiagonul matrix with Lii+l > 0. The mapping $ which maps L into the set {(A, -C X2 < . . . < A,), f: (f, f) = 1, fi > 0} is one to one. Furthermore, corresponding to any “spectral data” {(A, < A, < . . . <h,),f:(f,f)=l,fi‘>O}, one can associate a real symmetric tridiagonul matrix L in such a way that

(9 A,,A,,..., A, are the eigenvalues of L, (ii) the vector f is the first row of the matrix of the normalized eigenvec-

tars of L, and (iii) Li i + 1 > 0 for 1~ i < n - 1. In particularA?is homeomorphic to R”- I.


Proof. Suppose +(L,) = (p(L,). By Lemma 3 and remark (2) following that lemma, the vectors vi (L,) and vi (L,) are determined uniquely by f and A. Thus the matrix of eigenvectors is the same for both L, and L,. Hence L, = L,. In order to show that the map is onto, let vi, us,. . . , v, be the vectors obtained by applying the Gram-Schmidt process to the n vectors f, Af,..., A"-'f starting with f. The signs of the normalizations are chosen in such a way that ( vk, AkP 'f) > 0. This is possible because (ok, Akk- 'f) is not

zero (see the proof of Lemma 3). Set U = [ vl, vz, . . . , v,] t, and let L = UAU ‘. It follows from remark (1) made after Lemma 3 that L is tridiagonal, and by the choice of normalization (Lkpl),k = (v,, Akplf) > 0 for 2 < k < n. Hence by Lemma l(c), Li i + 1 > 0. As the set

is homeomorphic to lR ” - ‘, the second part of the theorem follows. W

Next an alternative choice of “spectral data” is given. Since Theorem 1 completely characterizes the matrices L, this alternative choice will be shown to be equivalent to the choice of Theorem 1.

DEFINITION. For n even let

/ a1 b, 0’ b, . .

s= b n/2%1

,O b n/Z-l a n/P

For n odd let

s=

i a1

bl

,O

b (+3x2

b (n-3)/2 a(,-3)/2 b(~A,,2

b (n-1)/2 a(.-,,/2


Sissaidtobeasubmatrixof~(P(X,,X,,...,X,)ifScanbecompletedtoan n X n tridiagonal matrix L which belongs to 9(X i, X a,. . . , A ,,). The space 9’of all submatrices is the new choice of “spectral data.”

We will show that the new data determines f with fi > 0, i = 1,2,. . . ,n, completely. Note that if the order of L is n, one needs R equations to determine ff,. . . ,f,“. It will be shown (Lemma 4) that the submatrix S determines ( Lj)ll for j = 1,. . . , n - 1 and conversely. Assuming this for the moment, it then follows that

n

(Lj),, = (Ajf,f)= c %.e (j=l,...,n-1) 77, = 1

and

2 f,"=l. m=l

Thus if

w = 1, L1l,L;l,...,L;;l)t (

then

where V[h,, A,,..., A,] is the Vandermonde matrix

v=

1 1 . . . 1

x, x2 .** x,

x2 . . . 1

x2 2

x2 R

jyy jy-’ . . . p-’

Conversely suppose o = (1, ti, E2,. . . ,[,_ 1)” is a vector such that

(V[h,, ha,..., X,]))‘w has a solution with alI elements positive. Then there is an L with (X1,X2,..., A,) and w [or equivalently the submatrix S-given by

Lemma 4-that corresponds to L,l,(L2)l,,(L3)11,...(Ln-1)ll with (Lj)ll = t,] as the spectral data. In order to see this, it is enough to let this positive solution be denoted by (f:, . . . , f,“) with fi > 0 and then recover L by Theorem 1. According to the theorem L has eigenvalues h,, x 2,. . . ,A “, a matrix of eigenvectors U with

f = U’e,,

51 DETERMINATION OF A TRIDIAGONAL MATRIX

and

We have thus proved

THEOREM 2. The space of submatrices of9(h1, A,, . . . ,A,) is the section of the cone

5=v[~~>~~,...J”lf, A‘>0

with the hyperplane 6. e, = 1.

Finally we must prove

LEMMA 4. Let L be a real symmetric tridiagonul n x n matrix with Li i+ I > 0. Let S be the submatrix corresponding to L. There is a one-one correspondence between the submatrix S and the elements

L,,,(L2),,,...,(L”_‘),,.

Proof. We prove the lemma when n = 2m is even. The odd case is similar. Let

L=

Then

s=

a1 h bl a2

0 b2

a1 bl

bl a2

b2

0

0

b2

b2

b 2mp1

0 \

b 2m-1

a2m J

0

b m-1

b m-1 anI


Let k < m. Then

(L2k-‘)ll = CLljlLj,j, ’ . . Ljk_,jkLjkjk+l . . . Lj2k_21> (*)

where the sum above is taken over j, = 1,2, jskP2 = 1,2, and If - j,_,l< 1 because L is tridiagonal. These restrictions imply that js can take values 1,2,3; j, can take values 1,2,3,4; and j,_ i can take values 1,2,. . . , k. (Note that Ljkm,jk is the central term in each expression on the right side of ( *) above, there being k - 1 terms on either side.) By the same argument jskP2 can take values 1,2; jsk_s can take values 1,2,3; and j, can take values 1,2,. . . , k. Thus

(LZk%i is completely determined by a,, b,,...,bk_l,ak. ln particular S determines L,,, ( L2),,, . . . , (L”-‘),,. Next observe that in the sum on the right side of ( * ) a k occurs only in one term, namely when jk _ i = k and jk = k. This necessarily forces j,,, = k - 1 = jkP2. Similarly jk+2 = k - 2 = j&3, etc. ThUS

the term involving uk contains no other a ‘s, and since Li i+ 1 z 0, it follows that if a,, b, ,..., u~_~, b,_, are determined by L,,,(L2),, ,..., (Lzkp2),,, then

a,, b i,...,uk are determined by L,,,(L2),,,...,(L2k-1)ll. Similarly one can show that a,, b,,. . . ,ak, b, are determined completely by L,,, (L2),,, . . . ,(L2k)11. Since L,, = a,, it follows by induction that

L,,,(L2),,,..., (L”- ‘)11 determines S completely. n

EXAMPLE. To illustrate the above ideas, consider hi < A, < A,. We will (i) describe what numbers a, and b, are allowed for the submatrix S, (ii) show how to obtain formulae for the remaining elements of the extension L.

(i) We have

u,=L,,= i xif;2, i=l

b: = (L’),, - uf =

= k$l ( xk- 5 Ail;‘)2ft* i=l

In other words, we see that the submatrix S = (u,b,) is allowable if and only if a, and b, are respectively the average and standard deviation of a


probability measure

WOO) i=l

concentrated on the points A, < A, < A,. (ii) Suppose a 1 and b, are respectively the average and standard deviation

of a probability measure p = Cf= lfi2S( A - A,), f; # 0. First we calculate u2.

We have

(L2),, = uf + by,

(L3),, = u,(af + b:)+ b,(a,b, + u,b,).

In particular

u2 = 3 [ ( L3),, - u; - 2a,bf]. 1

The eigenvalues A,, A,, A, satisfy

x3 - a,A2 + u2h - Is3 = 0,

where

u1 = A, + A, + A,,

a2 = X,h, + AJ, + X,A,,

a3 =x,x,x,.

By the Cayley-Hamilton theorem

uJ3)ll = 4% - %,~l, + 03

= ul( a; + bf ) - u2u, + a,.

Thus

U 2=~[ul(a~+b;)-u2u1+u3-u~-2u1b~]. 1


To calculate b,, consider

(L4)U = G3)rr+ b,[b,( u: + by) + Q,( a,b, + a&,) + b&l )

solve for b,, and use

as before. The calculation of a3 is similar, and the generalization to the n X n case is clear.

REMARK. The results of this paper do not extend in a straightforward way to higher order band matrices. For example, in the 3 X 3 case, consider the matrix

i

a1 bl Cl

L = b, u2 b,

Cl b2 a3

(cr > 0, spectrum = A) with associated submatrix

S = (u,b,c,).

Then, in contrast to the tridiagonal case, it turns out that the map

is generically two to one. The general n X n pentadiagonal case is, of course, more complicated. We plan to treat the problem in a later paper.

The authors would like to thank Professor Golub for bringing this problem to their attention and also they would like to thank the referee for pointing out an error in the original manuscript.

REFERENCES

1 H. Hochstadt, On the construction of a Jacobi matrix from mixed given data, Linear Algebra Appl. 28:113-115 (1979).

2 P. Deift, F. Lund, and E. Trubowitz, Nonlinear wave equations and constrained harmonic motion, Comm. Math. Phys. 74:141-188 (1980).


3 H. Hochstadt, On the construction of a Jacobi matrix from spectral data, Linear

Algebra A&. 8:435-446 (1974). 4 Ole H. Hald, Inverse eigenvalue problems for Jacobi matrices, Linear Algebra

Appl. 14:63-85 (1976). 5 H. Hochstadt and B. Lieberman, An inverse_Sturm Liouville problem with mixed

given data, SZAh4 J. A&. Math. 34:676-680 (1978). 6 G. H. Golub, unpublished results.

Receioed 20 july 1981; revised 28 January 1983

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